请教 []里如何区别开两个正则表达式

tasteoftime_90

举个例子：
[[[:space:]]*[[:lower:]]+[[:space:]]*\n]
[[:space:]]*[[:lower:]]+[[:space:]]*和\n区分开

谢啦！

tasteoftime_90

啊 对哦 忘了[]里面元字符无意义这块 就是想弄明白两行空白是哪里来的而已
[try@KeepTry patrol]$ cat try.txt
123 abc 456 def
abc 123 def 456
[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*"}{print $0}' try.txt
123
456

123
456

[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*"}{print RT}' try.txt
abc
def

abc
def

[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*";ORS="|"}{print $0}' try.txt
123|456||123|456
|[try@KeepTry patrol]$
[try@KeepTry patrol]$ awk 'BEGIN{RS="([[:space:]]*[[:lower:]]+[[:space:]]*|\n)"}{print $0}' try.txt
123
456

123
456

回复 2# bulletmarquis


   

tasteoftime_90

木有原始诉求 瞎弄 就是想弄明白两行空白是哪里来的而已
[try@KeepTry patrol]$ cat try.txt
123 abc 456 def
abc 123 def 456
[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*"}{print $0}' try.txt
123
456

123
456

[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*"}{print RT}' try.txt
abc
def

abc
def

[try@KeepTry patrol]$ awk 'BEGIN{RS="[[:space:]]*[[:lower:]]+[[:space:]]*";ORS="|"}{print $0}' try.txt
123|456||123|456
|[try@KeepTry patrol]$
[try@KeepTry patrol]$ awk 'BEGIN{RS="([[:space:]]*[[:lower:]]+[[:space:]]*|\n)"}{print $0}' try.txt
123
456

123
456
回复 4# yestreenstars


   

tasteoftime_90

我理解的是中间的空行是两个分隔符中间的“空”造成的，而下面的空行是“\n”造成的？
回复 10# bulletmarquis


   

tasteoftime_90

那么第一行后面没有\n 而第二行有 难道是以“def”结尾和以“456”结尾的不同？
回复 11# blackold


   

tasteoftime_90

明白了 多谢！
回复 14# blackold